BR physics Doppler Effect question

[anbuitachi]

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Hi this is a question from BR physics.

Submarine USA is moving at 10 m/s toward a second active sonar source (SASS). Both submarine USA and SASS emit same standard source freq. If Sub USA sends out a signal to SASS and then listens to echo, how would echo freq detected by Submarine USA compare to source frequency?

USA emits signals with 500 hz freq. Speed of sound in water is taken as 1500 m/s.

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I know that the echo freq> source due to Doppler shift but I don't understand the specifics of it. I dont understand the book explanation that well either.

They explained it as USA emit 500 hz and the first doppler shift brings it to 503 Hz using the formula f1 = f( v+v0 / v). Then as the echo is sent back, another doppler shift brings it to 507 Hz using the formula f2=f1 (v / v-vs), and 507 > 503 therefore echo > source.

I'm really confused about when to use double doppler effect regarding one moving object and 1 stationary. Can someone explain this to me please?

 

[joshto]

Hi this is a question from BR physics.

Submarine USA is moving at 10 m/s toward a second active sonar source (SASS). Both submarine USA and SASS emit same standard source freq. If Sub USA sends out a signal to SASS and then listens to echo, how would echo freq detected by Submarine USA compare to source frequency?

USA emits signals with 500 hz freq. Speed of sound in water is taken as 1500 m/s.

----
I know that the echo freq> source due to Doppler shift but I don't understand the specifics of it. I dont understand the book explanation that well either.

They explained it as USA emit 500 hz and the first doppler shift brings it to 503 Hz using the formula f1 = f( v+v0 / v). Then as the echo is sent back, another doppler shift brings it to 507 Hz using the formula f2=f1 (v / v-vs), and 507 > 503 therefore echo > source.

I'm really confused about when to use double doppler effect regarding one moving object and 1 stationary. Can someone explain this to me please?

i agree. got this one wrong. tough question.

 

[Geekchick921]

What you need to consider that they're asking you to consider the Doppler Effect twice. It's almost like in optics with a multiple lens system, where the image of the first lens serves as the object for the second. If the question were just what SASS observes from the USA sub, it would just be 503. But they're asking what frequency the USA sub detects on the echo.

As the USA sub moves toward the stationary SASS, the Doppler Effect causes a slight increase in frequency, and it goes from 500 to 503. 503 Hz is directed back towards the USA sub as the source frequency. Since the USA sub is still moving towards the SASS, the Doppler Effect causes another slight increase in frequency, up to 507.

I don't agree with this, though...

They explained it as USA emit 500 hz and the first doppler shift brings it to 503 Hz using the formula f1 = f( v+v0 / v). Then as the echo is sent back, another doppler shift brings it to 507 Hz using the formula f2=f1 (v / v-vs), and 507 > 503 therefore echo > source.

I think the first equation should be Fo = Fs (V / V-Vs). The second should be Fo = Fs (V+Vo / V). The only thing that's moving in this problem is the USA sub, which is the source in the first calculation, and the observer in the second.

 

[plsfoldthx]

can you write out the whole question please?>

oh nm i actually got this one right I just did

df/source f = v/c twice

 

[joshto]

What you need to consider that they're asking you to consider the Doppler Effect twice. It's almost like in optics with a multiple lens system, where the image of the first lens serves as the object for the second. If the question were just what SASS observes from the USA sub, it would just be 503. But they're asking what frequency the USA sub detects on the echo.

As the USA sub moves toward the stationary SASS, the Doppler Effect causes a slight increase in frequency, and it goes from 500 to 503. 503 Hz is directed back towards the USA sub as the source frequency. Since the USA sub is still moving towards the SASS, the Doppler Effect causes another slight increase in frequency, up to 507.

I don't agree with this, though...



I think the first equation should be Fo = Fs (V / V-Vs). The second should be Fo = Fs (V+Vo / V). The only thing that's moving in this problem is the USA sub, which is the source in the first calculation, and the observer in the second.

I just looked back, and it turns out I got this question right, not wrong. Regardless, in the explanation, it seems as though they have used Vs instead of Vo (in error) for the second doppler shift. Either way, the answer comes out to about the same value.

 

[anbuitachi]

What you need to consider that they're asking you to consider the Doppler Effect twice. It's almost like in optics with a multiple lens system, where the image of the first lens serves as the object for the second. If the question were just what SASS observes from the USA sub, it would just be 503. But they're asking what frequency the USA sub detects on the echo.

As the USA sub moves toward the stationary SASS, the Doppler Effect causes a slight increase in frequency, and it goes from 500 to 503. 503 Hz is directed back towards the USA sub as the source frequency. Since the USA sub is still moving towards the SASS, the Doppler Effect causes another slight increase in frequency, up to 507.

I don't agree with this, though...



I think the first equation should be Fo = Fs (V / V-Vs). The second should be Fo = Fs (V+Vo / V). The only thing that's moving in this problem is the USA sub, which is the source in the first calculation, and the observer in the second.

oh i think i get your point thanks

 

[dingyibvs]

Hi this is a question from BR physics.

Submarine USA is moving at 10 m/s toward a second active sonar source (SASS). Both submarine USA and SASS emit same standard source freq. If Sub USA sends out a signal to SASS and then listens to echo, how would echo freq detected by Submarine USA compare to source frequency?

USA emits signals with 500 hz freq. Speed of sound in water is taken as 1500 m/s.

----
I know that the echo freq> source due to Doppler shift but I don't understand the specifics of it. I dont understand the book explanation that well either.

They explained it as USA emit 500 hz and the first doppler shift brings it to 503 Hz using the formula f1 = f( v+v0 / v). Then as the echo is sent back, another doppler shift brings it to 507 Hz using the formula f2=f1 (v / v-vs), and 507 > 503 therefore echo > source.

I'm really confused about when to use double doppler effect regarding one moving object and 1 stationary. Can someone explain this to me please?

You use the double doppler effect when the emitter and the receiver are both moving. In this case, the sub is both the emitter and the receiver, and it is moving, therefore both the emitter and the receiver are/is moving.

 

[puravida85]

What you need to consider that they're asking you to consider the Doppler Effect twice. It's almost like in optics with a multiple lens system, where the image of the first lens serves as the object for the second. If the question were just what SASS observes from the USA sub, it would just be 503. But they're asking what frequency the USA sub detects on the echo.

As the USA sub moves toward the stationary SASS, the Doppler Effect causes a slight increase in frequency, and it goes from 500 to 503. 503 Hz is directed back towards the USA sub as the source frequency. Since the USA sub is still moving towards the SASS, the Doppler Effect causes another slight increase in frequency, up to 507.

I don't agree with this, though...



I think the first equation should be Fo = Fs (V / V-Vs). The second should be Fo = Fs (V+Vo / V). The only thing that's moving in this problem is the USA sub, which is the source in the first calculation, and the observer in the second.

I was thinking the exact same thing. If USA is moving towards the SASS and emits an echo, USA is the source and the SASS (we're not told its velocity) is the detector causing vD in V +/- vD to be 0. Which leads to Fo=Fs (V/V-VS) as you have written above. TBR isn't perfect after all. 🙂